By the way that is incorrect
Real Solution
Since the plants are indistinguishable, we must only count the number of plants on each window sill.
If 5 all the plants are on one window sill, there are ways to choose which window sill they are on.
If 5 plants are on one window sill and the last is on another, there are ways to choose which plants go on which window sill.
If 4 plants are on one window sill and the last two are on another, there are ways to choose which window sill they are on.
If 4 plants are on one window sill and the last two are each on one of the other windows, there are ways to choose which window the plants are on.
If 3 plants are on one window and the other plants are all on another window, there are ways to choose which window has no plants.
If 3 plants are on one window, plants on another window, and plant on the last window, there are ways to choose which plants are on which windows.
If 2 plants are on each window, there is only one way to arrange them.
In total, there are \(3+6+6+3+3+6+1=\boxed{28}\) ways to arrange the plants on the window sills.